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Let us say there is a container with two ideal gases in it divided by a piston.Volume of ideal below the piston is 2L and above is 8L.Total volume of container is 10L. n of both is 1 mole and T is 300 for both. Find the Temp when the ratio between the volume of gases is 4L and 6L.

How I solved it :

$\frac{R*1*300}{2L}$ - $\frac{R*1*300}{8L}$ since the gas below has more pressure.

Now my sir told me that difference in pressures is equal to weight of piston.The equation that I wrote above.My sir told me this because He said it is important into know that difference in pressures is equal to weight of piston .How is that possible? Please explain

Please do share if there is any other way to solve this if possible.

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  • $\begingroup$ The pressure in the lower chamber is equal to the pressure in the upper chamber plus the pressure exerted by the piston. This is because both the gas in the upper chamber, and the weight of the piston, are exerting a compressive force on the gas in the lower chamber. And the pressure exerted by the piston is the weight of the piston divided by its area. Your expression correctly calcuates the pressure exerted by the piston as the difference between the pressure in the upper chamber and the pressure in the lower chamber. $\endgroup$
    – theorist
    Nov 30 '20 at 11:42
  • $\begingroup$ The lower chamber gas is applying pressure on the upper chamber and piston. $\endgroup$
    – srijan Sri
    Nov 30 '20 at 11:51
  • $\begingroup$ You mean to say that P exerted by upper + piston = P by lower chamber. $\endgroup$
    – srijan Sri
    Nov 30 '20 at 11:52
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    $\begingroup$ Approach your tasks by algebraic solution first and plug in the numbers with units in the end. It has several benefits. 1/ Better grasp of underlying principles. 2/ Both of you and the readers can more easily spot a solution error or conceptual mistake. 3/ It gives to the Q/A the higher and more permanent value to be reused/applied for similar questions. 4/ It follows the site policy to focus on explaining principles and procedures, rather than solving of particular tasks with the one time value for 1 person only. 5/ It keeps you from intermediate rounding errors. $\endgroup$
    – Poutnik
    Nov 30 '20 at 11:53
  • $\begingroup$ Ok for sure from next time. $\endgroup$
    – srijan Sri
    Nov 30 '20 at 11:54
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We know from the first Newton motion law, that the net force acting on an object in rest must be zero. The forces acting on the piston are gravity and gas pressure:

$$\vec F_g + \vec F_\mathrm{p,down} + \vec F_\mathrm{p,up}=\vec 0 \tag{1}$$

If $V$ is the given bottom gas volume, $V_0$ is the total gas volume, $n$ is the molar amount of each of gases, the gas constant $R \approx \pu{8.314 JK^-1mol^-1}$, $T$ is absolute temperature. $m, g, A$ are piston mass, gravitational acceleration and piston area, then:

$$ m \cdot g + p_\mathrm{top} \cdot A =p_\mathrm{p,bottom}\cdot A\tag{2}$$

$$ p_\mathrm{bottom} - p_\mathrm{p,top} = \frac{m \cdot g}{A}\tag{3}$$

As we know from the ideal gas state equation that

$$p=\frac{nRT}{V}\tag{4}$$

then the pressure difference between the bottom and upper gases is

$$\Delta p = nRT ( \frac 1{V} - \frac 1{V_0-V})=\frac {mg}{A}\tag{5}$$

As the right side is constant, we can eliminate it, writing

$$nRT_1 ( \frac 1{V_1} - \frac 1{V_0-V_1}) = nRT_2 ( \frac 1{V_2} - \frac 1{V_0-V_2})\tag{6}$$

$$T_2 = T_1 \cdot \frac {\frac 1{V_1} - \frac 1{V_0-V_1}} {\frac 1{V_2} - \frac 1{V_0-V_2}}\tag{7}$$

where $T_1, V_1$ are the initial temperature and the volume of the bottom gas,$T_2, V_2$ are the final temperature and the volume of the bottom gas.

Now we finally plug in the numbers. Doing it before would be just distraction.

$$T_2 = \ce{300 K} \cdot \frac {\frac 1{\pu{2 L}} - \frac 1{\pu{8 L}}} {\frac 1{\pu{4 L}} - \frac 1{\pu{6 L}}} = \ce{300 K} \cdot \frac {3/8\ \pu{L^-1}} {1/12\ \pu{L^-1}}= \ce{300 K} \cdot \frac 92 = 1350 K\tag{8}$$

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